2005 AMC 12B Problems/Problem 10

Problem

The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous terms. What is the 2005th term of the sequence?

$\mathrm{(A)}\ {{{29}}} \qquad \mathrm{(B)}\ {{{55}}} \qquad \mathrm{(C)}\ {{{85}}} \qquad \mathrm{(D)}\ {{{133}}} \qquad \mathrm{(E)}\ {{{250}}}$

Solution

This problem is easily solved by induction. Now, performing this operation several times yields the results of 133 for the second term, 55 for the third term, and 250 for the fourth term. The sum of the cubes of the digits of 250 equal, 133, a complete cycle. The cycle is...excluding the first term, the 2nd, 3rd, and 4th terms will equal 133, 55, and 250, following the fourth term. Any multiple of three+1 as the term number will equal 250. It just so happens that 2005 is one more than a multiple of three, which leads us to the answer of 250.

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2005 AMC 12B Problems/Problem 10

Problem

Solution

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